If $\|g - g_x\|_{\infty}$ is small, then so is $\|g - g_x\|_p$

Question

Let $G$ be a locally compact (additive) abelian group with Haar measure $\mu$. Let $g \in C_c(G)$ with support $K$, and $1 \leq p < \infty$. Then $g$ is uniformly continuous on $G$, so there exists a neighborhood $V$ of the identity of $G$ such that $x, y \in G$ and $x- y \in V$ implies $|g(x) - g(y)| < \epsilon \mu(K)^{\frac{-1}{p}}$. Therefore, if $y$ is any element of $G$, and $x \in V$, then $y - (y-x) = x \in V$, and so $$|g(y) - g(y-x)| < \epsilon \mu(K)^{\frac{-1}{p}}$$ Therefore for each $x \in G$, we have $$\|g - g_x\|_\infty < \epsilon \mu(K)^{\frac{-1}{p}}$$ where $g_x$ is the function given by $g_x(y) = g(y-x)$. The book I'm reading (Rudin, Fourier Analysis on Groups) claims that as a result, we have $$\|g - g_x\|_p < \epsilon$$ but I'm not seeing it. I don't see where the support of $g$ is coming in here, and how I can relate it to the support of $g - g_x$.

Answer 1

I'm not sure if Rudin is wrong, but I was able to change the bound to make the proposition he's proving work out.

First, produce a neighborhood $V$ of the identity in $G$ so that $$||g - g_x||_{\infty} < \frac{\epsilon}{2} \mu(K)^{\frac{-1}{p}}$$ for all $x \in V$. If $y \not\in K \cup [K +x]$, then $g(y)$ and $g(y-x)$ are $0$. It follows that the support of $h := g - g_x$ is contained in $K \cup [K + x]$ (we just showed that $\{y \in G : h(y) \neq 0\}$ is contained in that union, hence the closure is also contained in there, since $K \cup [K + x]$) is compact. Therefore $$||g-g_x||_p^p = \int\limits_G |h|^p = \int\limits_{K \cup [K + x]} |h|^p \leq \int\limits_K |h|^p + \int\limits_{K+x} |h|^p \leq \frac{2 \epsilon^p}{2^p}$$ We just used the fact that $h$ vanishes off of $K \cup [K + x]$, and that the measures of $K$ and $K+x$ are the same. Therefore $||g - g_x||_p \leq \frac{\sqrt[p]{2}}{2} \epsilon \leq \epsilon$.